Q.Which of following solutions will have maximum boiling point elevation and which have minimum freezing point depression assuming the complete dissociation?
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Van't Hoff Factor
The Intuition: Why a Single Solute Can Act Like Many
Imagine you drop a sugar cube into a cup of tea. The sugar molecules stay whole — each molecule remains one particle. Now imagine you drop a pinch of table salt (NaCl) into water. The salt doesn't stay as NaCl molecules; it splits apart into Na⁺ and Cl⁻ ions. One formula unit of NaCl becomes two separate particles in solution.
If you measure the boiling point elevation or freezing point depression of these solutions, the salt solution behaves as if it has twice as many solute particles as the sugar solution — even though you dissolved the same number of formula units. That's the core idea: colligative properties depend on the number of particles, not the identity of the particles.
But what about substances that do the opposite? Acetic acid in benzene, for example, can pair up into dimers — two molecules associating to form one effective particle. That reduces the particle count.
The Van't Hoff factor i is simply the number that tells you: for every one formula unit you dissolve, how many particles actually end up floating around in solution?
The Precise Definition
i=expected value if no dissociation or association occursobserved value of a colligative property
For an ideal, non-dissociating, non-associating solute, i=1. For a solute that dissociates completely, i equals the number of ions produced per formula unit. For a solute that associates, i is less than 1.
How It Connects to Colligative Properties
Every colligative property formula gets multiplied by i:
- Relative lowering of vapour pressure: p∘p∘−p=i⋅n1n2
- Elevation of boiling point: ΔTb=i⋅Kb⋅m
- Depression of freezing point: ΔTf=i⋅Kf⋅m
- Osmotic pressure: Π=i⋅CRT
Where m is molality, C is molar concentration, and n1, n2 are moles of solvent and solute respectively.
The Two Cases in Detail
Dissociation (i > 1)
When a solute breaks into ions, the number of particles increases. For a compound AxBy that dissociates completely:
AxBy→xAy++yBx−
The maximum possible i is x+y. But if dissociation is partial, i lies between 1 and x+y.
The degree of dissociation α relates to i by:
i=1+(n−1)α
where n is the number of ions produced from one formula unit. For example, for NaCl (n=2), if α=0.8, then i=1+(2−1)(0.8)=1.8.
For complete dissociation of NaCl, i=2. For CaCl₂, i=3. For Al₂(SO₄)₃, i=5. Memorise these common ones — they appear frequently in numerical problems.
Association (i < 1)
When solute molecules combine to form larger aggregates, the particle count drops. For a dimerisation:
2A⇌A2
If n molecules associate to form one aggregate, and the degree of association is α:
i=1−(1−n1)α
For dimerisation (n=2), this becomes i=1−2α.
A common mistake: assuming i is always greater than 1. It is not. Association in non-polar solvents (like carboxylic acids in benzene) gives i<1. Always check whether the solute is likely to dissociate or associate in the given solvent.
A Worked Example …
Both effects scale with the total particle molality i × m, so counting dissociated particles per solution answers both parts at once. …
Particle tallies: KCl 0.2, NaCl 0.1, AlPO₄ 2.0, MgSO₄ 0.2 mol — so AlPO₄ has the maximum ΔTb and NaCl the minimum ΔTf.
Step 1. Boiling point elevation and freezing point depression are colligative: with complete dissociation, both are proportional to the TOTAL particle molality (i × m).
Step 2. Count the particles per kilogram of water for each option:
- (a) KCl → K⁺ + Cl⁻: 0.1×2=0.2 mol
- (b) NaCl → Na⁺ + Cl⁻: 0.05×2=0.1 mol
- (c) AlPO₄ → Al³⁺ + PO₄³⁻: 1×2=2.0 mol
- (d) MgSO₄ → Mg²⁺ + SO₄²⁻: 0.1×2=0.2 mol …
Multiply each molality by its ion count (all four salts here give 2 ions) and co …
- Assuming a salt with more highly charged ions automatically gives more particles — AlPO₄ still gives only 2 ions (Al³⁺ + PO₄³⁻); its high tally comes from its 1 m molality. …
Showing the 12 most recent of 22 on this concept.
- CBSE 2026Set V11 markQ.Van't Hoff factor(i) for a non-electrolyte in a solution is ________.
›Reveal solutionSolution
For a non-electrolyte solute the van't Hoff factor i=1, since the number of particles in solution is unchanged.
The van't Hoff factor is
i=number of formula units dissolvedactual number of particles in solution after dissociation/association …
- CBSE 2026Set ANNUAL1 markMCQQ.The strong electrolyte having maximum value of Van't Hoff factor(i) among the following is(a) NaCl(b) KCl(c) MgSO4(d) K2SO4
›Reveal solutionSolution
The van't Hoff factor i equals the number of ions a formula unit produces on complete dissociation, so the electrolyte giving the most ions has the largest i.
- NaCl -> Na+ + Cl- : i = 2
- KCl -> K+ + Cl- : i = 2
- MgSO4 -> Mg2+ + SO4^2- : i = 2
- K2SO4 -> 2K+ + SO4^2- : i = 3 …
- CBSE 2026Set ANNUAL1 markMCQQ.What is the value of Vant Hoff factor for one molal aqueous solution of Urea?(a) Two(b) Three(c) One(d) Four
›Reveal solutionSolution
Van't Hoff factor i measures how many particles a solute actually produces in solution. Urea is a non-electrolyte that neither dissociates nor associates, so i=1.
The Van't Hoff factor is defined as
i=calculated (normal) colligative propertyobserved colligative property=number of formula units dissolvedactual number of particles after dissolution
…
- CBSE 2026Set ANNUAL1 markMCQQ.When mercuric iodide is added to the aqueous solution of Potassium iodide:(a) Freezing point is raised(b) Freezing point does not change(c) Freezing point is lowered(d) Boiling point does not change
›Reveal solutionSolution
HgI2 combines with excess I− to form the complex ion [HgI4]2−, which decreases the number of particles in solution, so the depression in freezing point is less than expected and the freezing point is raised.
When mercuric iodide is added to an aqueous solution of potassium iodide, it reacts with the iodide ions already present:
HgI2+2I−→[HgI4]2− …
- CBSE 2026Set ANNUAL1 markQ.Write answer in one word/sentence: Write the numerical value of 'i' for complete dissociation of electrolyte K2SO4.
›Reveal solutionSolution
Complete dissociation of K2SO4 gives 3 ions, so i = 3.
The van't Hoff factor i is the number of particles actually present in solution per formula unit of the dissolved solute. On complete dissociation:
K2SO4 -> 2K+ + SO4^2-
…
- CBSE 2025Set ANNUAL1 markMCQQ.Which one of the following has maximum value of Van't Hoff Factor -(a) K2SO4(b) NaCl(c) MgSO4(d) KCl
›Reveal solutionSolution
Van't Hoff factor increases with the number of ions produced on complete dissociation.
For a strong electrolyte dissociating completely into n ions, the Van't Hoff factor i≈n.
- K2SO4→2K++SO42− : n=3
- NaCl→Na++Cl− : n=2
- MgSO4→Mg2++SO42− : n=2 …
- CBSE 2025Set ANNUAL1 markMCQQ.The value of Van't Hoff factor(i) for complete dissociation of MgSO4 is -(a) 1(b) 2(c) 3(d) 4
›Reveal solutionSolution
MgSO4 is a strong electrolyte that on complete dissociation gives 2 ions per formula unit, so i = 2.
The Van't Hoff factor is defined as:
i = (observed colligative property)/(calculated colligative property assuming no dissociation) = (actual number of particles in solution)/(number of formula units dissolved)
MgSO4 dissociates in water as:
MgSO4 -> Mg2+ + SO4^2-
One formula unit gives 2 ions (1 Mg2+ + 1 SO4^2-) on complete dissociation. So for complete dissociation: …
- CBSE 2025Set ANNUAL1 markMCQQ.For solutes which do not undergo any association or dissociation in solution, the Van't Hoff factor(i) will be –(i) less than 1(ii) more than 1(iii) equal to 1(iv) zero
›Reveal solutionSolution
The van't Hoff factor i measures the ratio of actual to normal number of particles; if a solute neither associates nor dissociates, that ratio is exactly 1.
The van't Hoff factor is defined as:
i=observed molar massnormal molar mass=number of particles as per formulaobserved number of particles after dissolution
- If the solute dissociates (e.g. NaCl → Na⁺ + Cl⁻), the number of particles increases, so i>1. …
- CBSE 2024Set ANNUAL1 markMCQQ.In comparison to a 0.01 M solution of glucose, the depression in the freezing point of 0.01 M MgCl2 solution is :(a) remains same(b) about twice(c) about three times(d) about four times
›Reveal solutionSolution
MgCl2 dissociates into 3 ions (van't Hoff factor i≈3), so its colligative effect — including freezing-point depression — is about 3 times that of a non-electrolyte like glucose at the same molar concentration.
Depression in freezing point is a colligative property proportional to the number of solute particles in solution:
ΔTf=i⋅Kf⋅m
Glucose is a non-electrolyte, so i=1 (it does not dissociate). …
- CBSE 2024Set ANNUAL1 markQ.What is Van't Hoff factor? OR Define Activation Energy.
›Reveal solutionSolution
Van't Hoff factor i measures how many particles a solute actually produces in solution compared to the 'ideal' assumption of one particle per formula unit.
Raoult's law and the colligative-property formulas (boiling-point elevation, freezing-point depression, osmotic pressure, etc.) assume the solute stays as single, non-interacting particles. But some solutes DISSOCIATE into more particles (e.g. NaCl → Na⁺ + Cl⁻, doubling the particle count) while others ASSOCIATE into fewer, larger particles (e.g. some carboxylic acids dimerise in non-polar solvents). This causes the OBSERVED colligative property to differ from the value calculated for a simple, non-dissociating solute.
The Van't Hoff factor is defined as:
i = (observed colligative property) / (calculated colligative property for no dissociation)
Equivalently, i = actual number of particles in solution after dissociation/association ÷ number of formula units initially dissolved.
- i > 1 → the solute dissociates (more particles than expected) …
- CBSE 2024Set ANNUAL1 markMCQQ.When benzoic acid is dissolved in benzene, the van't Hoff factor will be(a) 1(b) 0.5(c) 2(d) 1.5
›Reveal solutionSolution
Benzoic acid dimerises in benzene through intermolecular H-bonding, so the number of effective particles halves and i=1/n=0.5.
The van't Hoff factor i measures how the effective number of particles in solution deviates from the number of formula units dissolved:
i=observed (abnormal) molar massnormal molar mass=calculated colligative property (no association/dissociation)observed colligative property
In a polar, protic solvent (e.g. water), benzoic acid partially ionises, i>1. But benzene is non-polar, so ionisation cannot occur. Instead, two molecules of benzoic acid associate through two intermolecular hydrogen bonds between the −COOH groups of each molecule, forming a stable cyclic dimer:
2C6H5COOH⇌(C6H5COOH)2
If association is complete, every n=2 monomer units behave as a single kinetic particle, so
i=n1=21=0.5
…
- CBSE 2024Set ANNUAL1 markMCQQ.The values of Van't Hoff factors for KCl, NaCl, K2SO4 respectively are(a) 2, 2 and 2(b) 2, 2 and 3(c) 1, 1 and 2(d) 1, 1 and 1
›Reveal solutionSolution
Van't Hoff factor for a fully-dissociating strong electrolyte equals the number of ions it produces per formula unit.
KCl → K⁺ + Cl⁻ gives 2 ions ⟹ i = 2. NaCl → Na⁺ + Cl⁻ gives 2 ions ⟹ i = 2. K₂SO₄ → 2K⁺ + SO₄²⁻ gives 3 ions ⟹ i = 3. (These are the theoretical/limiting values assuming complete dissociation; measured i i …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.